Given this matrix:
\begin{matrix}
2 & 3 & 0 & 9 & 0 & 1 & 0 & 1 & 1 & 2 & 1 \\
1 & 1 & 0 & 3 & 0 & 0 & 0 & 9 & 2 & 3 & 1 \\
1 & 4 & 0 & 2 & 8 & 5 & 0 & 3 & 6 & 1 & 9 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 \\
2 & 2 & 4 & 1 & 1 & 2 & 1 & 6 & 9 & 0 & 7 \\
0 & 0 & 0 & 6 & 0 & 7 & 0 & 1 & 0 & 0 & 0 \\
2 & 5 & 0 & 7 & 0 & 4 & 6 & 8 & 5 & 1 & 3 \\
0 & 0 & 0 & 1 & 0 & 4 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 8 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 2 & 1 & 1 \\
2 & 6 & 0 & 1 & 0 & 30 & 0 & 2 & 3 & 2 & 1 \\
\end{matrix}
Is there an efficient way of calculating its determinant using the matrix minors? it just seems that there's lots of zeroes inside so it has to ease up the operation and I'm missing something.
Thanks.
Best Answer
It is possible to notice that the third column has only one non-zero value, $4$. Hence we cancel out the third column and the fifth row. Now the (original) fourth row has only one non-zero value, $5$, hence we cancel out the fourth row and the sixth column and continue this way. After six steps, i.e. after using the elements in positions $(5,3),(4,6),(3,5),(7,7),(9,4),(8,8)$ as "pivots", we can notice that no non-zero element on the sixth row remains, so the determinant is zero.